Quiz 3
1. Simplify!
2x + 5
4
−
x2 + 2x + 1 2x + 2
6x
2x3 + 4x2 + 2x
Answer. Pick a place to start, and begin simplifying piece by piece. Lets
start with the top. Factoring the bottom of the first term gies (x + 1)2 , and
the bottom of the second term factors to 2(x + 1), and so the second term
4
2
is 2(x+1)
= x+1
.
Now, the top looks like :
2x + 5
2
−
.
2
(x + 1)
x+1
We need to find a common denominator to combine these fractions. The
LCM of the two denominators is (x + 1)2 , so we multiply the second one
by x+1
x+1 . This leads to...
x2
2x + 5
2
x+1
2x + 5
2(x + 1)
−
·
=
−
2
+ 2x + 1 x + 1 x + 1
(x + 1)
(x + 1)2
(2x + 5) − 2(x + 1)
=
(x + 1)2
2x + 5 − 2x − 2
=
(x + 1)2
3
.
=
(x + 1)2
So we have simplified the entire top of the fraction to
3
(x+1)2 .
Now lets work on the bottom. We can factor a 2x out of the denominator
to get 2x(x2 + 2x + 1) = 2x(x + 1)2 . But then this 2x cancels with hte 6x
on top to leave a 3 on top. Therefore the entire bottom of the fraction is
3
(x+1)2 .
So now that we’ve simplified the top and the bottom, the expression
looks like:
3
(x+1)2
3
(x+1)2
1
.
This simplifies to
3
3
(x + 1)2
3
÷
=
·
= 1.
(x + 1)2
(x + 1)2
(x + 1)2
3
2. Simplify!
1
1
1
+
+
x2 − x − 6 x2 + x − 2 x2 − 4x + 3
Answer. Step 1: Factor all of the denominators. This gives
1
1
1
+
+
(x − 3)(x + 2) (x + 2)(x − 1) (x − 3)(x − 1)
Now, to find the LCM, we look at all of the factors of each denominator.
(x + 2) is raised to the 1 in the first and second term, but zero in the third,
so it is raised to the one in the LCM. Repeating this for the other two, we
get that
LCM = (x − 3)(x + 2)(x − 1). Now we make all of the denominators equal
by multiplying each term on top and bottom by the appropriate factor.
x−1
x−3
x+2
+
+
(x − 1)(x + 2)(x − 3) (x − 1)(x + 2)(x − 3) (x − 1)(x + 2)(x − 3)
Now that the denominators are equal, we just add across the numerators.
3x − 2
(x − 1) + (x − 3) + (x + 2)
=
(x − 1)(x + 2)(x − 3)
(x − 1)(x + 2)(x − 3)
3. True of False?
1
1
+
=0
x−7 7−x
b) The polynomial x2 − x + 2 is prime
a)
Answer. a) True. Since 7 − x = −(x − 7), the left side becomes
1
1
1
1
+
=
−
= 0.
x − 7 −(x − 7)
x−7 x−7
b) True. There are no two factors of 2 which add up to −1.
2